What is the integral of (3-2x+x^2) ?

The integral of (3-2x+x^2) is 3x-x^2+(x^3)/3+C

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integral of (3-2x+x^2)

\int (3 - 2 x + x^{2}) d x

3 x - x^{2} + \frac{x ^{3}}{3} + C

Solution steps

\int 3 - 2 x + x^{2} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 3 d x - \int 2 x d x + \int x^{2} d x

\int 3 d x = 3 x

\int 2 x d x = x^{2}

\int x^{2} d x = \frac{x ^{3}}{3}

= 3 x - x^{2} + \frac{x ^{3}}{3}

Add a constant to the solution

= 3 x - x^{2} + \frac{x ^{3}}{3} + C

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What is the integral of (3-2x+x^2) ?
The integral of (3-2x+x^2) is 3x-x^2+(x^3)/3+C